Solveeit Logo
Login

Question

Question: 500 ml, 1M NaCl(aq.) solution is mixed with 1000 ml, 1M $AgNO_3$(aq.) solution. Which following opti...

500 ml, 1M NaCl(aq.) solution is mixed with 1000 ml, 1M AgNO3AgNO_3(aq.) solution. Which following option is correct for resultant solution :

A

[Na+]=13M[Na^+] = \frac{1}{3}M

B

[Ag+]=23M[Ag^+] = \frac{2}{3}M

C

[NO3−]=43M[NO_3^-] = \frac{4}{3}M

D

[Cl−]=13M[Cl^-] = \frac{1}{3}M

Answer

A

Explanation

Solution

1. Identify the Reaction and Initial Moles:

The reaction between NaCl(aq) and AgNO₃(aq) is a precipitation reaction:

NaCl(aq)+AgNO3(aq)→AgCl(s)+NaNO3(aq)NaCl(aq) + AgNO_3(aq) \rightarrow AgCl(s) + NaNO_3(aq)

Calculate the initial moles of each reactant:

  • Moles of NaCl:

    Volume = 500 mL = 0.5 L Molarity = 1 M Moles of NaCl = Molarity × Volume = 1 mol/L × 0.5 L = 0.5 mol

    This means we have 0.5 mol of Na+Na^+ ions and 0.5 mol of Cl−Cl^- ions initially.

  • Moles of AgNO₃:

    Volume = 1000 mL = 1.0 L Molarity = 1 M Moles of AgNO₃ = Molarity × Volume = 1 mol/L × 1.0 L = 1.0 mol

    This means we have 1.0 mol of Ag+Ag^+ ions and 1.0 mol of NO3−NO_3^- ions initially.

2. Determine Limiting Reactant and Moles After Reaction:

The precipitation reaction involves Ag+Ag^+ and Cl−Cl^- ions:

Ag+(aq)+Cl−(aq)→AgCl(s)Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)

  • We have 1.0 mol of Ag+Ag^+ and 0.5 mol of Cl−Cl^-.
  • Since the stoichiometry is 1:1, Cl−Cl^- is the limiting reactant.
  • 0.5 mol of Cl−Cl^- will react completely with 0.5 mol of Ag+Ag^+ to form 0.5 mol of AgCl(s).

Now, let's find the moles of each ion after the reaction:

  • Moles of Na+Na^+: Na+Na^+ is a spectator ion. It does not participate in the reaction.

    Initial moles of Na+Na^+ = 0.5 mol. Final moles of Na+Na^+ = 0.5 mol.

  • Moles of Cl−Cl^-: Cl−Cl^- is the limiting reactant and precipitates as AgCl.

    Initial moles of Cl−Cl^- = 0.5 mol. Moles of Cl−Cl^- reacted = 0.5 mol. Final moles of Cl−Cl^- ≈ 0 mol (assuming complete precipitation, as AgCl is highly insoluble).

  • Moles of Ag+Ag^+: Ag+Ag^+ is in excess.

    Initial moles of Ag+Ag^+ = 1.0 mol. Moles of Ag+Ag^+ reacted = 0.5 mol. Final moles of Ag+Ag^+ = 1.0 mol - 0.5 mol = 0.5 mol.

  • Moles of NO3−NO_3^-: NO3−NO_3^- is a spectator ion. It does not participate in the reaction.

    Initial moles of NO3−NO_3^- = 1.0 mol. Final moles of NO3−NO_3^- = 1.0 mol.

3. Calculate Total Volume of the Resultant Solution:

Total Volume = Volume of NaCl solution + Volume of AgNO3AgNO_3 solution Total Volume = 500 mL + 1000 mL = 1500 mL = 1.5 L

4. Calculate Final Concentrations of Ions:

  • [Na+][Na^+]: [Na+]=Moles of Na+Total Volume=0.5 mol1.5 L=13 M[Na^+] = \frac{\text{Moles of Na}^+}{\text{Total Volume}} = \frac{0.5 \text{ mol}}{1.5 \text{ L}} = \frac{1}{3} \text{ M}

  • [Ag+][Ag^+]: [Ag+]=Moles of Ag+Total Volume=0.5 mol1.5 L=13 M[Ag^+] = \frac{\text{Moles of Ag}^+}{\text{Total Volume}} = \frac{0.5 \text{ mol}}{1.5 \text{ L}} = \frac{1}{3} \text{ M}

  • [NO3−][NO_3^-]: [NO3−]=Moles of NO3−Total Volume=1.0 mol1.5 L=23 M[NO_3^-] = \frac{\text{Moles of NO}_3^-}{\text{Total Volume}} = \frac{1.0 \text{ mol}}{1.5 \text{ L}} = \frac{2}{3} \text{ M}

  • [Cl−][Cl^-]: [Cl−]=Moles of Cl−Total Volume=0 mol1.5 L≈0 M[Cl^-] = \frac{\text{Moles of Cl}^-}{\text{Total Volume}} = \frac{0 \text{ mol}}{1.5 \text{ L}} \approx 0 \text{ M}

5. Compare with Options:

(A) [Na+]=13M[Na^+] = \frac{1}{3}M (Matches our calculation) (B) [Ag+]=23M[Ag^+] = \frac{2}{3}M (Our calculation: 13M\frac{1}{3}M) (C) [NO3−]=43M[NO_3^-] = \frac{4}{3}M (Our calculation: 23M\frac{2}{3}M) (D) [Cl−]=13M[Cl^-] = \frac{1}{3}M (Our calculation: ≈0M\approx 0M)

Therefore, option (A) is correct.